航空发动机强度振动上机作业题题目三班级：140411姓名：***学号：********一：题目要求3-1某转子叶片根部固定，其材料密度ρ=2850kg/m3,弹性模量E=71.54GPa，叶片长0.1m,各截面位置、面积、惯性矩列于下表，试求其前3阶固有静频。
二：分析公式如题目所示，已知材料密度，弹性模量，各截面位置、面积、惯性矩,需要求解3阶静频率。
求解静频公式如下：1.弹性线归一化2.振型逼近法3.二阶振型和固有频率求解4.三阶振型和固有频率求解三：编程计算程序使用c语言编写，源代码如下：#include<stdio.h>#include<math.h>#include<stdlib.h>int main(void){float rou=2850;float E=71540000000;float X[11]={0.0,0.01,0.02,0.03,0.04,0.05,0.06,0.07,0.08,0.09,0.10};floatA[11]={0.00017,0.000146,0.000126,0.000109,0.000096,0.000086,0.000077,0.000073 ,0.00007,0.000068,0.000068};floatI[11]={0.000000000279,0.000000000212,0.000000000157,0.000000000108,0.00000 0000084,0.000000000061,0.000000000045,0.000000000037,0.000000000032,0.000 000000030,0.000000000030};float Ab[10];float Ib[10];int i=0;while(i<=9){Ab[i]=((A[i+1]+A[i])/2);i=i+1;}i=0;while(i<=9){Ib[i]=((I[i+1]+I[i])/2);i=i+1;}float Y0js[10]={0,0.1,0.2,0.4,0.5,0.6,0.7,0.8,0.9,1};float Y0sj[10]={0,0,0,0,0,0,0,0,0,0};float Y0xxx[10]={0,0,0,0,0,0,0,0,0,0};float one[10]={0,0,0,0,0,0,0,0,0,0};float two[10]={0,0,0,0,0,0,0,0,0,0};float three[10]={0,0,0,0,0,0,0,0,0,0};float four[10]={0,0,0,0,0,0,0,0,0,0};float wucha=10;float xz=2.58;int a=0;int b=0;int c=0;int d=0;int e=0;int f=0;int g=0;int h=0;int j=0;int k=0;int p=0;int q=0;while(wucha>=0.000001) //给定拟合精确度{q=0;while(9>=q){one[q]=0;two[q]=0;three[q]=0;four[q]=0;q=q+1;} //这步是给数组清零，千万不能忘！a=0;b=9;while(9>=a){b=a;while(9>=b+1){one[a]=one[a]+(Ab[b+1]*Y0js[b+1]*0.01);b=b+1;}a=a+1;} //第一重循环c=0;d=0;while(9>=c){d=c;while(9>=d+1){two[c]=two[c]+(one[d+1]*0.01);d=d+1;}c=c+1;} //第二重循环e=0;f=0;while(9>=e){f=0;while(e>=f){three[e]=three[e]+two[f]*0.01*(1/Ib[f]);f=f+1;}e=e+1;} //第三重循环g=0;h=0;while(9>=g){h=0;while(g>=h){four[g]=four[g]+three[h]*0.01;h=h+1;}g=g+1;} //第四重循环k=0;while(9>=k){Y0sj[k]=four[k]/four[9];Y0xxx[k]=four[k];k=k+1;} //求出实际y0，以便和假设yo对比迭代wucha=0;j=0;while(j<=9){wucha=wucha+fabs(Y0js[j]-Y0sj[j]);j=j+1;} //假设的y0与求出y0之间的误差p=0;while(9>=p){Y0js[p]=Y0sj[p];p=p+1;} //令实际值等于假设值，再次迭代运算}float omega=0;omega=5000*sqrt((1/four[9]));printf("一阶固有静频为：%.5fHZ\n",omega/(2*3.1415926));float Y2js[10]={0,-0.1,-0.3,-0.5,-0.4,-0.2,-0.1,0.5,0.8,1};float Y2[10]={0,0,0,0,0,0,0,0,0,0};float xiuzhen=4;float Y2sj[10]={0,0,0,0,0,0,0,0,0,0};float b11=0;float C21=0;float a21=0;a21=10;while(fabs(a21)>=0.00000001) //给定拟合精确度{b11=0;C21=0;a21=0;i=0;while(i<=9){b11=b11+2850*Ab[i]*Y0xxx[i]*Y0xxx[i]*0.01;i++;}i=0;while(i<=9){C21=C21+2850*Ab[i]*Y0xxx[i]*Y2js[i]*0.01;i++;}a21=C21/b11;i=0;while(i<=9){Y2[i]=Y2js[i]-a21*Y0xxx[i];i++;}p=0;while(9>=p){Y2js[p]=Y2[p];p=p+1;} //令实际值等于假设值，再次迭代运算}i=0;while (i<=9){Y2sj[i]=Y2[i]/Y2[9];i++;} //归一化q=0;while(9>=q){one[q]=0;two[q]=0;three[q]=0;four[q]=0;q=q+1;} //这步是给数组清零，千万不能忘！a=0;b=0;while(9>=a){b=a;while(9>=b+2){one[a]=one[a]+(Ab[b+2]*Y2sj[b+2]*0.01);b=b+1;。
